In the figure (1) given below, ABCD is a parallelogram in which ∠DAB = 70°, ∠DBC = 80°. Calculate angles CDB and ADB.

As AD || BC, sum of co-int ∠s = 180°.

∠A + ∠B = 180°

∠B = 180° - ∠A = 180° - 70° = 110°.

From figure,

⇒ ∠B = ∠DBA + ∠DBC

⇒ 110° = ∠DBA + 80°

⇒ ∠DBA = 110° - 80° = 30°.

Sum of angles of a triangle = 180°

In △DAB,

⇒ ∠DAB + ∠DBA + ∠ADB = 180°

⇒ 70° + 30° + ∠ADB = 180°

⇒ ∠ADB = 180° - 100° = 80°.

Since, opposite angles of a parallelogram are equal,

∴ ∠C = ∠A = 70°.

In △DBC,

⇒ ∠DBC + ∠BCD + ∠CDB = 180°

⇒ 80° + 70° + ∠CDB = 180°

⇒ ∠CDB = 180° - 150° = 30°.

Hence, ∠ADB = 80° and ∠CDB = 30°.